I'm working on differential manifolds, and at a moment in my lesson there is a pull-back of a vector field on an open sets of ${\mathbb R}^n$ to a differential manifold $M$ by a function from charts of $M$. Do that make any sence ? Because this functions are homeomorphism, and not diffeomorphism, I'm a bit confused.
Asked
Active
Viewed 19 times
0
-
For a topological manifold, the charts are only homeomorphisms. Once you have equipped your manifold with a differentiable structure, the charts actually turn into differentiable maps. – Sven Pistre Sep 16 '22 at 12:26
-
Ah okay, it is because all the change of charts are diffeomorphisms ? – Johny06 Sep 16 '22 at 12:36
-
Exactly. See my answer. – Sven Pistre Sep 16 '22 at 12:38
1 Answers
0
To expand on my comment: A-priorily, charts are only homeomorphisms onto their image.
By definition, a map $F\colon M\to N$ between two smooth manifolds $M$ and $N$ is smooth if and only if for every point $p\in M$ there is a chart $(U,\phi)$ for $M$ containing $p$ and a chart $(V,\psi)$ for $N$ with $F(U)\subset V$ such that $\psi\circ F\circ \phi^{-1}$ is smooth from $\phi(U)$ to $\psi(V)$.
In your case, the map $F$ we want to check smoothness for is a fixed chart $\theta\colon U\subset M \to \theta(U)\subset \mathbb{R}^{m}$ where $U$ and $\theta(U)$ are open subsets (and thus submanifolds themselves) of $M$ and $\mathbb{R}^m$. The standard smooth atlas on $\mathbb{R}^{m}$ is given by $\{(\mathbb{R}^m, \operatorname{Id})\}$.
Consider any point $p$ and a chart $(U,\phi)$ for $M$ containing this point. Then we need to see if the following map is smooth: $$ \operatorname{Id} \circ \theta \circ \phi^{-1} = \theta \circ \phi^{-1} $$ But this map is just the transition map between the charts $\phi$ and $\theta$ and is therefore smooth by definition of the smooth structure on $M$! Note that you can do the same for $\theta^{-1}$.
So, a-posteriorily, charts on a smooth manifold are diffeomorphisms onto their image.
Sven Pistre
- 1,320